Answer:
Liquids evaporate faster as they heat up and more particles have enough energy to break away. The particles need energy to overcome the attractions between them. ... At this point the liquid is boiling and turning to gas. The particles in the gas are the same as they were in the liquid they just have more energy.
Why do we use semiconductor instead of metal in thermopile?
Answer:
Metal or conductors , what they do is that they allow full flow of current that is conduction is due to free electrons only and there is literally no gap between valence and conduction band,so free electrons can easily jump into conduction band from valence band.
And now lets talk about insulators,what they do is that they don’t allow any current to flow i.e they act as strong dielectric,and gap between valence band and conduction band is so big that free electron can never come into conduction band from valence band ever if they try is for eternity…
But semiconductors can act both as an insulator as well as a conductor based on the voltage input. Hence, there is a possibility to control the current flow in semiconductors , so they don’t just relax and let the current pass by, they can control it, and that is why you can design logic circuits with it.
But as the temperature increases free electrons from valence band of insulators can jump to conduction band and can cause a little conductivity, and then the insulator will act as a semi-conductor.
can you guys pls also solve for average speed.
Answer:
d_t = 3.05km
v_a = 4.3km/h
Explanation:
42mins*(2/3) = 28mins
42mins-28mins = 14mins
d = v*t
d_1 = (4km/h)*(1h/60mins)*(28mins)
d_1 = 1.87km
d_2 = (5km/h)*(1h/60mins)*(14mins)
d_2 = 1.17km
d_t = d_1+d_2
d_t = 1.87km+1.17km
d_t = 3.05km
v_a = (v_1+v_2)/2
v_a = [(2*4km/h)+5km/h)]/3
v_a = 4.3km/h
A 50.0 ohm and a 30.0 ohm resistor are connected in parallel. What is their equivalent resistance? Unit=Ohms
R(parallel) = product/ sum
50×30/50+30
1500/80
18,75 ohms
Answer: 18.75
above is right but you need to put a dot after the number 18
A plastic box with objects has a mass of 4 kg and is on a shelf at a height of 2.4 m. What will it's potential gravitational energy?
Answer:
potential energy=mgh
4×9.8×2.4
Explanation:
may be hope this will help you
2. The vector sums of and the Ark witar must se rue our the directions and maintedes at an Bit CB? What meast le tue about the directions and magnitudes and it cor
Check attached photo
Check attached photo
A/An is a type of blood cell that's also called a red blood cell. a) Jeukocyte O b) thrombocyte c) plasma d) erythrocyte
Answer:
red blood cell, also called erythrocyte
Explanation:
Hope it helps
Mark me as Brainliest plz
:DD
calculate the rate of loss of heat through a glass window of area 200 CM square and thickness 0.5 CM where temperature inside is 35 degree Celsius and outside is -5 degree Celsius conductivity of Glass is 2.2 into 10 to the power 3 cal per s per cm per k .
Answer:
The inner and outer surfaces of a 0.5-cm thick 2-m by 2-m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass
Explanation:
Flag question
Consider the pressure and force acting on the
dam retaining a reservoir of water. Suppose the
dam is 500-m wide
and the water is 80.0-m
deep at the dam, as illustrated below. What is
the average pressure on the dam due to the
water?
Answer:
P = density (p) * g * h
P = 1000 kg/m^3 * 9.8 m/s^2 * 40 m = 392,000 N/m^2
since kg m / s^2 = Newtons
The average pressure is 1/2 (pressure at 0m + pressure 80 m) for liquid of uniform density
7) A force of 500N exists between two identical point charges separated by a dis-
tance of 40 cm the magnitude of the two charges is
Answer:
q=9.43×10^-5C
Explanation:
F=kq^2/r^2
500= 9×10^9 × q^2/ (0.4)^2
500N×0.16m=9×10^9Nm^2C^2 × q^2
80/(9×10^9)= q^2
√(8.8889×10^9) = q
q= 9.43×10^-5C
since they are identical both charges, q=9.43×10^-5C
The magnitude of the two charges is 9.43×10^-5 C.
To calculate the magnitude of an electric force, it is necessary to use the following expression:
[tex]F = k \frac{q_{1} \times q_{2} }{d^{2}}[/tex]
Assuming that the constant is:
[tex]k = 9\times 10^{9}[/tex]
We can apply the values in the formula above, obtaining:
[tex]500 = 9 \times 10^{9} \times \frac{q^{2}}{0.4^{2}}[/tex]
[tex]q = \sqrt{0.88\times10^{-8}}[/tex]
[tex]q = 9.43 \times 10^{-5} C[/tex]
So, the magnitude of the two charges is 9.43×10^-5 C.
Learn more about electrical force in: brainly.com/question/16888648
10. A mass weighing 10 pounds stretches a spring foot. This mass is removed and replaced with a mass of 1.6 slugs, which is initially released from a point foot above the equilibrium position with a downward velocity of . (a) Express the equation of motion in the form given in (6). (b) Express the equation of motion in the form given in (6) (c) Use one of the solutions obtained in parts (a) and (b) to determine the times the mass attains a displacement below the equilibrium position numerically equal to the amplitude of motion.
Answer:
Hah yyytttdvooophhfdbbb hhhctuionmm hhdtjjjov
Current is the rate at which charge is flowing.
a. True
b. Fals
Answer:
A. True
Explanation:
Which item will be shipped third?
—-
Answer:
I know it's groceries
Explanation:
electronics ship before clothing
electronics ship after groceries
urgent items are first so
order:
1.) A/Electronics
2.) Clothing/B
3.) Groceries(since groceries aren't urgent)
thing is it's C or D I'm leaning to D since it says it ships last but i dont know so if I'm wrong sorry.
. A patient on the infectious disease floor takes 10 mL of levofloxacin syrup bid. If the product is only
available as a 5 mL unit-dose oral syringe, how many syringes will the technician prepare for a 24-
hour supply?
Answer:
The technician will need to prepare 48 syringes for a 24 hour supply since the patient needs double the available dose.
Explanation:
brainliest plz . . .
The pressure of sea water increases by 1.0atm for each 10m increase in the depth, by what percentage is the density of water increased in the deepest ocean of water of 12km. Compressibility is 5.0×10^-5 atm
The percentage by which the water density increased is 4.1[tex]\mathbf{\overline 6}[/tex] %
The known values are;
The increase in pressure per 10 meter increase in depth = 1.0 atm
The depth of the deepest ocean = 12 km = 12,000 m
The compressibility of the ocean = 5.0 × 10⁻⁵ 1/atm
The unknown
The percentage the density of water increased in the deepest ocean
Strategy;
Find the pressure at the deepest point of the deepest ocean and apply the compressibility
We have;
[tex]\mathbf{Compressibility = \dfrac{1}{V} \times \dfrac{\partial V}{\partial p}}[/tex]
The change in pressure, [tex]\partial p[/tex] = (12,000 m/(10 m)) × 1.0 atm = 1,200 atm
Therefore, we have for one cubic meter of water
[tex]\mathbf{5.0 \times 10^{-5} \ atm^{-1} = \dfrac{1}{1 \, m^3} \times \dfrac{\partial V}{1,200 \, atm}}[/tex]
Therefore;
[tex]\mathbf{\partial}[/tex]V = 5.0 × 10⁻⁵ atm⁻¹ × 1 m³ × 1,200 atm = 0.06 m³
The new volume = V - [tex]\mathbf{\partial}[/tex]V
∴ The new volume = 1 m³ - 0.06 m³ = 0.94 m³
The initial density = mass/(1 m³)
The new density = mass/(0.96 m³)
The percentage increase in density, [tex]\partial[/tex]ρ%, is given as follows;
[tex]\mathbf{\partial p \% = \dfrac{ \dfrac{Mass}{0.96 \ m^3} - \dfrac{Mass}{1 \ m^3} }{ \dfrac{Mass}{1 \ m^3}} \times 100 = \dfrac{25}{6} \% = 4.1 \overline 6 \%}[/tex]
∴ [tex]\mathbf{\partial}[/tex]ρ% = 4.1[tex]\mathbf {\overline 6}[/tex] %
The percentage by which the water density increased, [tex]\partial[/tex]ρ% = 4.1[tex]\mathbf{\overline 6}[/tex] %
Learn more about compressibility here;
https://brainly.com/question/18746977
What would happen if you changed the position of the screen, but kept the other factors the same?
Answer:
I wish I could help but iam srry
Our system is a block attached to a horizontal spring on a frictionless table. The spring has a spring constant of 4.0 N/m and a rest length of 1.0 m, and the block has a mass of 0.25 kg.
Compute the PE when the spring is compressed by 0.50 m.
Answer
E - 1/2 K x^2 potential energy of compressed spring
E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m
A stone that is dropped freely from rest traveled half the total height in the last second. with what velocity will it strike the ground
Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
A liquid is poured into a vessel to a depth of 16cm when viewed from the top, the bottom appears to be raised 4cm. What is the refractive index of the liquid?
Answer:
Solution
Verified by Toppr
Correct option is
C
3 cm
RI=apparent depthreal depth
Substituting, 34=apparentdepth12
Therefore, apparent depth=412×3=9
The height by which it appears to be raised is 12−9=3cm
Was this answer helpful?
71
0
SIMILAR QUESTIONS
A coin is placed at the bottom of a glass tumbler and then water is added. It appeared that the depth of the coin has been reduced because
Medium
View solution
>
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B
[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
Explanation:
Given:
[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]
[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by
[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]
[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]
[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
A 2.5 m segment of wire supplying current to the motor of a submerged submarine carries 1050 A and feels a 3.6 N repulsive force from a parallel wire 5.00 cm away. show answer No Attempt What is the magnitude of the current in the other wire in A
force between two parallel wire is
f/l = mueo*i1i2/2pir
f/l = 2*10^-7*i1i2/rl
i2 = f*r/2*10^-7*i1
i2 = 342.9 A
The kinetic energy of a particle of mass 500g is 4.8j. Determine the velocity of the particle
Answer:
4.38 m/s
Explanation:
What is the submarine's maximum safe depth in sea water? The pressure inside the submarine is maintained at 1.0 atm
Answer:
The submarine's maximum safe depth in sea water is 801.678 m.
Explanation:
P=Po+(rho)*g*h
Max Pressure = Initial Pressure + (Water Density)(Gravity)(Max Depth)
Area of Window = Pi*(Diameter/2)^2 = Pi*(.4m/2)^2 = 0.125664 m^2
Max Pressure= (1.0*10^6 N)/(0.125664 m^2)= 7.95775-E6 Pa
Initial Pressure= 1atm= 101.3kPa= 101300Pa
Water Density (rho) = 1000kg/m^3
Gravity= 9.8m/s^2
So rearranging for h= (P-Po)/((rho)*g)
h=((7.95775-E6Pa)-(101300Pa))/((1000kg/m^3)(9.8m/s^2))= 801.678 m
Help please!!!!!!!!! I will mark brainliest!!!
Answer:
solving for: velocity
equation: velocity = distance / time
substitution: velocity = 1425 km / 12.5 hrs
answer: 114 km/hr
A 4kg block is attached to a vertical spring with a spring constant of 800N/m. How much elastic potential energy is stored in the system?
E= [tex]\frac{1}{2}[/tex]×k×x² = 1J
given k=800n/m x=0.05m
What is a benefit of joining the YMCA instead of a private health club?
The YMCA has nicer equipment and buildings.The YMCA has nicer equipment and buildings. , ,
The YMCA is likely to be in an upscale neighborhood.The YMCA is likely to be in an upscale neighborhood. , ,
The YMCA is much more affordable.The YMCA is much more affordable. , ,
The YMCA has restaurants and spa services.
Answer:
Explanation:
This depends on where you are.
The YMCA where I grew up was affordable. Very. Or my mother would not have let me go.
1) Un objeto realiza un movimiento circular uniforme en una circunferencia de 10 metros de diámetro y efectúa 20 vueltas por minuto. Se pide hallar:
a) El periodo.
b) La frecuencia en Hertz.
c) La velocidad tangencial. d) La velocidad angular.
e) La aceleración centrípeta.
Answer:
A RECIPE NEEDS A COMBINED WEIGHT OF 720G OF FLOUR AND SUGAR. IF THE RECIPE NEEDS 5TIME FLOUR THAN SUGAR,HOW MUCH OF EACH IS NEEDED
If an object with constant mass is accelerating, what does Newton's second
law imply?
A. It will continue to accelerate until it meets an opposing force.
B. The object is exerting an opposite but equal force.
C. A force must be acting on the object.
D. The object will be difficult to decelerate.
Answer:
C. A force must be acting on the object.
Explanation:
This is due to the action of its momentum direction.
[tex].[/tex]
14. If you are asked to analyze an issue for an essay assignment, what should you do? a. Nb. C. Divide the issue into its main parts and discuss each part. Express your opinion about the issue. Explain what you think the issue means and how you came to that interpretation Describe the main ideas or points about the issue. d.
Answer:
Option C
Explanation:
Divide the issue into its main parts and discuss each part.
A student graphs power (p) on the vertical axis and time (t) on the horizontal axis. The graph appears to be a hyperbola.
a) What should the student graph on each axis to test whether the relationship is actually
hyperbolic?
b) If the relationship is actually hyperbolic, what is the general equation for the relationship between power and time?
Answer: it would be daddy
Explanation:
Because I’m daddy
A horizontal spring with spring constant 85 N/m extends outward from a wall just above floor level. A 2.5 kg box sliding across a frictionless floor hits the end of the spring and compresses it 6.5 cm before the spring expands and shoots the box back out. How fast was the box going when it hit the spring
As the box compresses the spring, the spring performs
-1/2 (85 N/m) (0.065 m)² ≈ -0.18 J
of work on the box. By the work energy theorem, the total work performed on the box (which is done only by the spring since there's no friction) is equal to the change in the box's kinetic energy. At full compression, the box has zero instantaneous speed, so
W = ∆K ==> -0.18 J = 0 - 1/2 (2.5 kg) v ²
where v is the box's speed when it first comes into contact with the spring. Solve for v :
v ² ≈ 0.14 m²/s² ==> v ≈ 0.38 m/s
